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WINSTONCH [101]
3 years ago
14

How far would a spacecraft moving in a circular orbit 500 km above Pluto's surface travel during that time?

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

The speed of the spacecraft should be 719.35m/s

Explanation:

if the spacecraft is orbiting the planet with a circular orbit, the gravitational force must act as a centripetal force. This means:

F_G=F_c\\\frac{GmM}{d^2}=m\frac{v^2}{d}

In this case, the pluto's mass M is 1.3099·10^22 kg. The radius of the planet R is 1188.3Km and G is the gravitational constant. Therefore:

\displaystyle\frac{G M}{d^2}=\frac{v^2}{d}\\v=\sqrt{\frac{GM}{d} } =\sqrt{\frac{GM}{(R + 500km)} } =719.35m/s

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The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

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E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

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3 years ago
The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball
Fudgin [204]

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

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How many grains are there in the ball?

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